Answer:
Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.
The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
Answer:
The correct answer is B. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K
Explanation:
Iron is hotter and gold is colder, therefore, according to laws of thermodynamics, iron will lose heat to gold until they are at the same temperature.
The specific heat capacity of iron(0.449) is over three times that of gold(0.128). Since masses are equal, this means that each time iron's temperature drops by one degree, the energy released it releases makes gold's temperature increase by more than 3 degrees. So gold's temperature will be climbing much faster than iron's is falling. Meaning they will meet closer to the initial temperature of iron than that of gold
Answer:
ΔH = 
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
=
where
= the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
= the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = + 0
==> ΔH =
Explanation:
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